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How many zeros does `f(x)=x^3-3x^2+a` have on [-1,1] with `a in RR` ?
(1) First notice that `a` creates a vertical translation of the graph of `y=x^3-3x^2` .
(2) If you are using calculus, you can find the local maximum and minimum by taking the first derivative and setting `f'=0` . Thus
Without calculus, you can inspect the graph for the turning points which occur at x=0 and x=2.
(3) Varying the value of `a` shifts the graph vertically. When `a=0` we have two real roots (The zero at x=0 is a double root). Then by inspecting the graphs we find:
When `a=-1` we have one real zero.
When `a=1` we have 3 real zeros.
-- Using calculus you find that the function increases on `(-oo,0)` ; has a relative maximum at x=0, decreases on (0,2); has a relative minimum at x=2; and increases on `(2,oo)` using the first derivative test.
Then finding that for `f(x)=x^3-3x^2` the relative maximum occurs at (0,0) you can conclude a double root. Shifting the function down yields a single real root. Find ing the relative minimum for `f(x)=x^3-3x^2` occurs at (2,-4) you realize a vertical shift up 4 results in a double root, and any greater vertical shift returns to a single root. Between 0 and 4 the function has 3 real roots --
Thus, the number of real zeros on [-1,1] depends on the value of a:
`-1<=a<0` On this interval the function has 1 real zero.
`a=0` At a=0 the function has two real zeros; one is a repeated root
`0<a<=1` On this interval the function has 3 real roots.
(( A complete characterization of the zeros:
For a<0 there is 1 real zero
For a=0 there are two real zeros -- one is repeated
For 0<a<4 there are 3 real zeros
For a=4 there are two real zeros -- one is repeated
for a>4 there is 1 real zero))
a=-1 black;a=0 red;a=2 green;a=4 blue;a=5 purple (Notice a is the y-intercept)
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